Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{-4n^2 + 36n}{n^3 - 13n^2 + 36n} \div \dfrac{-3n - 6}{-5n^2 + 25n - 20} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4n^2 + 36n}{n^3 - 13n^2 + 36n} \times \dfrac{-5n^2 + 25n - 20}{-3n - 6} $ First factor out any common factors. $q = \dfrac{-4n(n - 9)}{n(n^2 - 13n + 36)} \times \dfrac{-5(n^2 - 5n + 4)}{-3(n + 2)} $ Then factor the quadratic expressions. $q = \dfrac {-4n(n - 9)} {n(n - 4)(n - 9)} \times \dfrac {-5(n - 4)(n - 1)} {-3(n + 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-4n(n - 9) \times -5(n - 4)(n - 1) } { n(n - 4)(n - 9) \times -3(n + 2)} $ $q = \dfrac {20n(n - 4)(n - 1)(n - 9)} {-3n(n - 4)(n - 9)(n + 2)} $ Notice that $(n - 4)$ and $(n - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {20n\cancel{(n - 4)}(n - 1)(n - 9)} {-3n\cancel{(n - 4)}(n - 9)(n + 2)} $ We are dividing by $n - 4$ , so $n - 4 \neq 0$ Therefore, $n \neq 4$ $q = \dfrac {20n\cancel{(n - 4)}(n - 1)\cancel{(n - 9)}} {-3n\cancel{(n - 4)}\cancel{(n - 9)}(n + 2)} $ We are dividing by $n - 9$ , so $n - 9 \neq 0$ Therefore, $n \neq 9$ $q = \dfrac {20n(n - 1)} {-3n(n + 2)} $ $ q = \dfrac{-20(n - 1)}{3(n + 2)}; n \neq 4; n \neq 9 $